Leetcode 46 Full Permutation

difficulty ： secondary
The frequency of ：119
subject ：
Give an array without duplicate numbers nums , Back to its All possible permutations . You can In any order Return to the answer .

Their thinking ： Backtracking traversal

class Solution {
    
    public List<List<Integer>> permute(int[] nums) {
    
         // The length of the array 、 Number of children 
         int len=nums.length;
         // Create the final result  res
         List<List<Integer>> res=new ArrayList<>();
         // If the array length is 0, Then there is no need to arrange 
         if(len==0) return res;
        // take path Put it in deque in , Because Deque You can take out the front and back 
         Deque<Integer> path=new ArrayDeque<>(len);
         // Of the tag flag, If this number is used , Mark for 1, Not used yet 0 
         boolean[] flag=new  boolean[len];
         //DFS Traverse   It's actually doing LEN Time DFS Traverse 
         DFS(nums,len,0,path,flag,res);
         return  res;
    }

    public void DFS(int[] nums,int len,int depth,Deque<Integer> path, boolean[] flag
                                    , List<List<Integer>> res){
    
        
        if(len==depth){
    
            res.add(new ArrayList<>(path));
            return ;
        }
        //len The second traversal is highlighted here  ||  And one. DFS Trees len Three elements are also highlighted here 
        for(int i=0;i<len;i++){
    
            // If flag yes 0, That is, this number has not been used yet 
            if(!flag[i]){
    
                // If you haven't used it, use this number as follows 
                path.addLast(nums[i]);// The root node 
                flag[i]=true;
                DFS(nums,len,depth+1,path,flag,res);
                flag[i]=false;
                path.removeLast();
            }
        }

    }
}